# Problems in Multivariabe Calculus

Hello,

Ive got slight problems which i need to solve and understand for a entry test for a course which i hope to gain entrance into this week and im stuck on two questions on example test which i need help. see attached file can you please help me.

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#### Solution Preview

6. (a) We have f(x) = x^(2/3) - 2 x^(1/3) for x >= 0.

i. The extrema occur at the points (x, f(x)) in which f'(x) = 0. We have

f'(x) = 2/3 x^(-1/3) - 2/3 x^(-2/3)

= 2/3 x^(-2/3) (x^(1/3) - 1),

so f'(x) = 0 implies x = 1. We have f(1) = 1 - 2 = -1, so the point (1, -1) is an extremum. Since there are no other values of x for which f'(x) = 0, it is the only extremum. To see what type of extermum it is, we compute f''(1). We have

f''(x) = -2/9 x^(-4/3) + 4/9 x^(-5/3)

so f''(1) = -2/9 + 4/9 = 2/9 > 0, implying that (1, -1) is a local minimum.

ii. The points of inflection (x, f(x)) of the graph of f occur at the values of x in which f''(x) = 0. We have

f''(x) = 2/9 (2 - x^(1/3))

so f''(x) implies x^(1/3) = 2, which implies x = 8. We have

f(8) = 8^(2/3) - 2*8^(1/3)

= 4 - 2*2 = 0

so (8, 0) is an inflection point. Since there are no other values of x for which f''(x) = 0, it is the only inflection point.

(b) Let x(t) be the distance from the base of the ladder to the corner of the room and let y(t) be the distance from the top of the ladder to the corner of the room. Then we ...

#### Solution Summary

We solve various problems in multivariable calculus. The real valued functions on its largest domain and vectors are provided.